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11.1 Midpoint and Distance Formulas
11.2 Circles
11.3 Ellipses
11.4 Hyperbolas
11.5 Parabolas
11.6 Identifying Conic Sections
11.7 Solve Quadratic Systems
Conic Sections
Chapter 11
11.1 Midpoint and Distance Formulas
The midpoint of a line segment is the point equidistant from the
endpoints of the line segment. If you know the coordinates of the end-
points of a line segment, you can find the coordinates of the midpoint
by averaging the coordinates.
Midpoint Formula
The midpoint of a line segment with
endpoints (x1
,y1
) and (x2
,y2
) is the
point .
Example: Find the coordinates of the midpoint of line segment AB if
A(–4, 3) and B(6, –1).
Solution: Use the midpoint formula.
A(–4, 3), B(6, –1)
= (1, 1)
The midpoint of is (1, 1).
If you know the midpoint and one endpoint of a segment, you can use
the midpoint formula to calculate the other endpoint.
Example: The midpoint of line segment CD is (4, –2) and C (2, 3).
What are the coordinates of D?
Solution: Write equations for the x and y coordinates of D using
the midpoint formula and the coordinates of the midpoint. Solve
for x and y.
M = (4, –2), C(2, 3)
+ +⎛⎝⎜
⎞⎠⎟
x x y y2
,
2
1 2 1 2
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
= − + + −⎛⎝⎜
⎞⎠⎟
M 4 6
2
,
3 ( 1)
2
AB
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
Master Math: Algebra 22
(x1, y1)
(x2, y2)
( ),x1 + x2
2
y1 + y2
2
Conic Sections
Write equations for x2
and y2
from the
midpoint formula.
Multiply both sides of each equation by 2.
Solve.
The coordinates of D are (6, –7).
To find the distance between two points, you can use the distance for-mula, which is derived from the Pythagorean Theorem. Recall that the
Pythagorean Theorem says that the sum of the squares of the legs of a
right triangle is equal to the square of the hypotenuse, or .
Distance Formula
The distance between two points (x1
,y1
) and (x2
,y2
) is
.
Example: Find the length of line segment AB if A(–4, 3) and B(6, –1).
Solution: Use the distance formula.
A(–4, 3), B(6, –1)
Distance formula.
Substitute known values.
=+ x
4
2
2
2 − =+ y
2
3
2
2
= + x8 2
2
− = + y4 3
2
= x6
2
− = y7
2
= +c a b2 2 2
= +c a b2 2 2
( ) ( )= − + −d x x y y2
2 1
2
2 1
2
( ) ( )= − + −d x x y y2 1
2
2 1
2
= − + −d x x y y2
2 1
2
2 1
2
( ) ( )= − + −d x x y y2 1
2
2 1
2
( ) ( )= − + −d x x y y2 1
2
2 1
2
− =+ +⎛
⎝⎜⎞⎠⎟
x y(4, 2)
2
2
,
3
2
2 2
( ) ( )= − − + − −d 6 ( 4) 1 3
2 2
3
(x1, y1)
|x2 – x1|
|y2 – y1|d
(x1, y2)(x2, y2)
Simplify.
Simplify.
The length of is approximately 10.77 units.
Practice Exercises
11.1 Find the coordinates of the midpoint of line segment AB if A(1, 7)
and B(–5, 3).
11.2 The midpoint of line segment CD is (–3, 1) and endpoint
C is (2, 5). What are the coordinates of D?
11.3 Find the length of line segment AB if A(–2, –5) and B(8, –4).
11.2 Circles
A circle is the set of points in a plane that
are equidistant from a single point, called
the center of the circle. The radius of the
circle is a line segment whose endpoints
are the center of the circle and any point
on the circle.
Since a circle is defined by a distance from
a point, you can use the distance formula
to write the equation of a circle with center
(h, k).
Distance formula.
Substitute values.
Square both sides.
Standard Form of the Equation of a Circle
Center at (h, k) and radius r Center (0, 0) and radius r
= ≈d 116 10.77
AB
( ) ( )− + − =x x y y d2 1
2
2 1
2
− + − =x h y k r( ) ( )
2 2
− + − =x h y k r( ) ( )
2 2 2
( )= + −d 10 4
2
2
+ =x y r2 2 2− + − =x h y k r( ) ( )
2 2 2
Master Math: Algebra 24
(x, y)
(h, k)
Conic Sections
You can think of the graph of a circle with center (h, k) as a transformation
of the circle with center (0, 0). Every point on the circle is translated
h units horizontally and k units vertically.
Example: Graph . Is the circle a function?
Solution: Identify the center and the radius of the circle. Graph
the center and sketch the circle using the radius.
Standard equation of a circle.
Given equation.
Transform to standard form.
Center (–1, 3), radius 2
The equation of a circle is not
a function. You can see in the
graph that there are two y values
for most x values. The circle
fails the vertical line test for a
function.
Example: Write the equation of a circle with center (6, –5) and radius .
Solution: Substitute h, k, and r in the standard form of the equation
of a circle.
h = 6, k = –5, r =
Standard equation of a circle.
Substitute h, k, and r.
Simplify.
+ + − =x y( 1) ( 3) 4
2 2
− + − =x h y k r( ) ( )
2 2 2
+ + − =x y( 1) ( 3) 4
2 2
− − + − =x y( ( 1)) ( 3) 2
2 2 2
7
7
− + − =x h y k r( ) ( )
2 2 2
− + − − =x y( 6) ( ( 5)) 7
2 2
2
− + + =x y( 6) ( 5) 7
2 2
5
-4 -3 -2 -1 1
2
3
1
5
4
(x + 1)2 + (y – 3)2 = 4
(-1, 3)
(-1, 5)
(-1, 1)
(-3, 3) (1, 3)
Note that the center of a circle is a reference point and not part of the circle.The only points on the circle are the points that are r units away from thecenter. That includes the points marked with dots as well as the infinitelymany points that lie on the graph of the circle.
The diameter of a circle is a line segment that passes through the center
of a circle and has endpoints on the circle. You can write the equation for
a circle if you know the endpoints of the diameter by using the distance
and midpoint formulas.
Example: Write the equation of a circle if the endpoints of a diameter
are (2, 5) and (–6, 3).
Solution: Use the midpoint formula to find the center of the circle.
Use the distance formula to find the radius. Substitute h, k, and rin the standard form of the equation of a circle.
Midpoint formula.
Substitute endpoints of diameter.
(h, k) = (–2, 4) Simplify.
The center of the circle is (–2, 4).
Find the distance between the center and another point on the circle.
Distance formula.
Substitute center (–2, 4) and point (2, 5).
Simplify.
Simplify.
The radius of the circle is .
Use the center and the radius to write the equation of the circle.
Standard equation of a circle.
Substitute h, k, and r.
Simplify.
If you are given the equation of a circle that is not in standard form,
you may have to complete the square in order to transform the equation
to standard form. To review completing the square, visit Lesson 5.3.
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
= + − +⎛⎝⎜
⎞⎠⎟
h k( , )
2 6
2
,
5 3
2
( ) ( )= − + −d x x y y2 1
2
2 1
2
( ) ( )= − − + −r 2 2 4 5
2 2
= +r 16 1
=r 17
17
− + − =x h y k r( ) ( )
2 2 2
− − + − =x y( ( 2)) ( 4) 17
2 2
2
+ + − =x y( 2) ( 4) 17
2 2
Master Math: Algebra 26
Conic Sections 7
Example: Graph the circle .
Solution: Transform the equation into the standard form for the
equation of a circle by completing the square for x and y. Identify
the center and the radius and sketch the graph.
Regroup.
.
Add 13 to both sides.
Factor the perfect square
trinomials.
Write in standard form.
Center (2, –3), radius 6 Identify the center and radius.
+ − + =x y x y4 6 23
2 2
+ − + =x y x y4 6 23
2 2
− + + =x x y y( 4 ) ( 6 ) 23
2 2
− + + + + = +x x y y( 4 4) ( 6 9) 23 13
2 2
−⎛⎝⎜
⎞⎠⎟
=⎛⎝⎜
⎞⎠⎟
=4
2
4;
6
2
9
2 2
− + + =x y( 2) ( 3) 36
2 2
− + + =x y( 2) ( 3) 6
2 2 2
-4 -3 -2 -1 1 2 3 4 5 6 7 8
2
3
1
-1
-2
-3
-4
-5
-6
-7
-8
-9
4
(x – 2)2 + (y + 3)2 = 36
(2, -3)
(2, 3)
(2, -9)
(-4, -3) (8, -3)
Practice Exercises
11.4 Graph .
11.5 Write the equation of a circle with center (2, 1) and radius 9.
11.6 Write the equation of a circle if the endpoints of the diameter
are (–1, –4) and (3, –8).
11.7 Write the equation of the circle in
standard form.
11.3 Ellipses
An ellipse looks like a circle that has been flattened. An ellipse is the set
of all points in the plane such that the sum of the distances from two fixed
points is constant. Each fixed point is called a focus, or together, the foci.
An ellipse has two axes of symmetry: the longer one called the majoraxis and the shorter one called the minor axis. The axes are perpendicu-
lar and intersect at the midpoint of the foci, or the center of the ellipse.
The foci lie on the major axis.
An ellipse has four vertices. Each vertex is the point of intersection of
the ellipse and an axis of symmetry. Some call only the intersections
of the major axis and the ellipse vertices and call the intersection of the
minor axis and the ellipse minor vertices or co-vertices.
Ellipse with horizontal major axis Ellipse with vertical major axis
+ + + = −x y x y12 18 106
2 2
+ + + =x y( 5) ( 2) 1
2 2
Master Math: Algebra 28
co-vertex
co-vertex
center
major axis vertexvertex
minor axis
focusfocus
co-vertexco-vertex center
major axis
vertex
vertex
minor axis
focus
focus
Conic Sections 9
Standard Form of the Equation of an Ellipse Centered at (0, 0)
Major Axis Horizontal Vertical
Equation , a > b , a > b
Foci (–c, 0), (c, 0) (0, –c), (0, c)
a, b, cLength of major axis 2a 2a
Length of minor axis 2b 2b
Vertices (–a, 0), (a, 0) (0, –a), (0, a)
Co-vertices (0, –b), (0, b) (–b, 0), (b, 0)
You can write the equation of an ellipse when you are given the coordi-
nates of the vertices and the foci.
Example: Write the equation of the ellipse shown on the next page.
Solution: The ellipse has a horizontal major axis. Find the length
of the major axis and divide by 2 to find a. Calculate a2
. Identify
c from the foci, and use c2
= a2
– b2
to find b2
. Substitute a2
and b2
into the standard form of the equation of an ellipse with a horizontal
major axis.
+ =xb
ya
1
2
2
2
2
+ =xa
yb
1
2
2
2
2
= −c a b2 2 2= −c a b2 2 2
There are two forms of the standard equation of an ellipse centered at
(0, 0), depending on whether the major axis is horizontal or vertical.
When an ellipse is written in standard form, you can identify the direction of the major axis by identifying which variable has the larger denominator. If x2 has a larger denominator, the major axis will be the x-axis or parallel tothe x-axis. If y2 has a larger denominator, the major axis will be the y-axis or parallel to the y-axis.
If a = b, then the major and minor axes are equal, and the ellipse is a circle.
The coordinates of the vertices that lie on the major axis are (–7, 0)
and (7, 0), so the length of the major axis is 14.
2a = 14 Length of the major axis
a = 7 Divide both sides of the equation by 2.
a2
= 49 Calculate a2
.
The foci are (–6, 0) and (6, 0), so c is 6.
c2
= a2
– b2
Relationship among a, b, and c.6
2
= 49 – b2
Substitute c = 6 and a2
= 49.
36 = 49 – b2
Simplify.
–13 = –b2
Subtract 49 from both sides of the equation.
13 = b2
Multiply both sides of the equation by –1.
Standard equation of ellipse with a horizontal
major axis.
Substitute a2
= 49 and b2
= 13.
You can write the equation of an ellipse centered at the origin if you
know the lengths of both axes.
+ =xa
yb
1
2
2
2
2
+ =x y49 13
1
2 2
Master Math: Algebra 210
-4 -3 -2 -1 1 2 3 4 5 6 7 8-5-6-7-8
-1
-2
-3
-4
3
4
2
1
(6, 0) (7, 0)(-7, 0) (-6, 0)
Conic Sections
Example: If the length of the horizontal axis of an ellipse is 20 and the
length of the vertical axis is 16, write the equation of the ellipse
centered at the origin in standard form and graph the ellipse.
Solution: Since the horizontal axis is longer, it is the major axis.
The length of the major axis is 2a, and the length of the minor axis
is 2b. Set each expression equal to the length of each axis and find
the value of a and b. Substitute a and b into c2
= a2
– b2
and solve
to find the value of c. Substitute a and b in the standard form of the
equation of an ellipse with a horizontal major axis. Identify and
graph the foci, vertices, and co-vertices. Sketch a graph of the ellipse.
2a = 20 2b = 16 c2
= a2
– b2
a = 10 b = 8 c2
= 10
2
– 8
2
c2
= 36
c = 6
Standard equation of ellipse with a horizontal
major axis.
Substitute a = 10 and b = 8.
Simplify.
+ =xa
yb
1
2
2
2
2
+ =x y10 8
1
2
2
2
2
+ =x y100 64
1
2 2
11
-2 42 10 126 8-4-6-8-10-12
-2
-4
-6
-8
6
8
4
2
(-10, 0) (-6, 0) (10, 0)(6, 0)
(0, 8)
(0, -8)
+ = 1x2
100y2
64
Foci: (–c, 0), (c, 0) → (–6, 0), (6, 0)
Vertices: (–a, 0), (a, 0) → (–10, 0), (10, 0)
Co-Vertices: (0, –b), (0, b) → (0, –8), (0, 8)
Notice that in standard form, the equation for an ellipse is equal to 1.
You may have to transform an equation into standard form.
Example: Write the equation in standard form and sketch
a graph of the ellipse.
Solution: Divide both sides of the equation by 63. Identify a2
and
b2
and solve to find a and b. Use the equation c2
= a2
– b2
to find c.
Plot the foci and vertices and sketch the graph.
Divide both sides of the equation by 63.
Since the denominator of y2
is larger, the ellipse has a vertical
major axis and the standard equation is .
+ =x y7 3 63
2 2
+ =x y7 3 63
2 2
+ =x y9 21
1
2 2
+ =xb
ya
1
2
2
2
2
Master Math: Algebra 212
-1 21 3 4-2-3-4
-1
-2
-3
-4
-5
3
4
5
2
1
+ = 1x2
9y2
21
(-3, 0) (3, 0)
(0, 3.46)
(0, -3.46)
(0, -4.58)
(0, 4.58)
Conic Sections
c2
= a2
– b2
b = 3 c2
= 21 – 9
a ≈ 4.58 c2
= 12
c ≈ 3.46
Foci: (0, –c), (0, c) → (0, –3.46), (0, 3.46)
Vertices: (0, –a), (0, a) → (0, –4.58), (0, 4.58)
Co-Vertices: (–b, 0), (b, 0) → (–3, 0), (3, 0)
The graph of an ellipse can be translated so that the center is (h, k). Just
as you have seen with other translated graphs, x – h and y – k replace xand y.
Standard Form of the Equation of an Ellipse Centered at (h, k)
Major Axis Horizontal Vertical
Equation , a > b , a > b
Foci (h – c, k), (h + c, k) (h, k – c), (h, k + c)
a, b, c c2
= a2
– b2 c2
= a2
– b2
Length of 2a 2amajor axis
Length of 2b 2bminor axis
Vertices (h – a, k), (h + a, k) (h, k – a), (h, k + a)
Co-vertices (h, k – b), (h, k + b) (h – b, k), (h + b, k)
If the equation of an ellipse is not in standard form, you may have to
complete the square in order to transform the equation to standard form.
− + − =x hb
y ka
( ) ( )
1
2
2
2
2
− + − =x ha
y kb
( ) ( )
1
2
2
2
2
=a 21
2 =b 9
2
=a 21
13
Example: Write the equation in standard
form and sketch a graph of the ellipse.
Solution: Transform the equation into standard form by completing
the square for x and y, and then divide both sides of the equation by
a value that will make the equation equal to 1. Identify a2
and b2
and solve to find a and b. Use the equation c2
= a2
– b2
to find c.
Plot the foci and vertices and sketch the graph.
Regroup and factor.
Factor.
Divide both sides by 16.
Since the denominator of x2
is larger, the ellipse has a horizontal
major axis and the standard equation is .
The center of the ellipse is (5, –3).
− + + + + = − +x x y y2( 10 25) 8( 6 9) 106 122
2 2
⎛⎝⎜
⎞⎠⎟
=⎛⎝⎜
⎞⎠⎟
=2
10
2
50, 8
6
2
72
2 2
+ − + = −x y x y2 8 20 48 106
2 2
− + + = −x x y y2( 10 ) 8( 6 ) 106
2 2
+ − + = −x y x y2 8 20 48 106
2 2
− + + =x y2( 5) 8( 3) 16
2 2
− + + =x y( 5)
8
( 3)
2
1
2 2
− + − =x ha
y kb
( ) ( )
1
2
2
2
2
Master Math: Algebra 214
2 54 86 731-1-2-3
-1
-2
-3
-4
1+ = 1x2
8y2
2
(5, -3)
(5, -1.59)
(5, -4.41)
(7.83, -3)(2.55, -3)(2.17, -3)
(7.45, -3)
+ = 1(x – 5)2
8(y + 3)2
2
Conic Sections
a2
= 8 b2
= 2 c2
= a2
– b2
b2
= 2 c2
= 8 – 2
a ≈ 2.83 b ≈ 1.41 c2
= 6
c ≈ 2.45
Foci: (h – c, k), (h + c, k)
(5 – 2.45, –3), (5 + 2.45, –3)
(2.55, –3), (7.45, –3)
Vertices: (h – a, k), (h + a, k)
(5 – 2.83, –3), (5 + 2.83, –3)
(2.17, –3), (7.83, –3)
Co-Vertices: (h, k – b), (h, k + b)
(5, –3 – 1.41), (5, –3 + 1.41)
(5, –4.41), (5, –1.59)
Notice in the previous example that you could also graph the ellipse by
translating the graph of down 3 and to the right 5.
You can also write the equation of an ellipse if you know the coordinates
of the vertices and the co-vertices.
Example: Write the equation of the ellipse in standard form if the ver-
tices of the ellipse are (2, 19) and (2, –7) and the co-vertices
are (–3, 6), and (7, 6).
Solution: Since the vertices have the same x coordinate, the major
axis is vertical. Find the center of the ellipse by finding the mid-
point of the major or minor axis. Find the length of the major and
minor axes. The length of the major axis is 2a, and the length of
the minor axis is 2b. Set each expression equal to the length of each
axis and find the value of a and b. Substitute a and b in the standard
form of the equation of an ellipse with a vertical major axis.
+ =x y8 2
1
2 2
=a 8
15
Master Math: Algebra 216
Midpoint formula.
Substitute the coordinates of the vertices.
M = (2, 6)
The center of the ellipse is (2, 6). Or, h = 2 and k = 6.
Length of the vertical axis = 19 – (–7) = 26.
Length of the horizontal axis = 7 – (–3) = 10.
2a = 26 2b = 10
a = 13 b = 5
Standard equation of ellipse with vertical
major axis.
Substitute known values.
Simplify.
Practice Exercises
11.8 Write the equation of the ellipse on the next page.
11.9 If the length of the horizontal axis is 8 and the length of the
vertical axis is 10, what is the equation of the ellipse centered
at the origin?
11.10 Write the equation in standard form and sketch the
graph of the ellipse.
11.11 Write the equation 11 in standard form
and sketch the graph of the ellipse.
11.12 Write the equation of the ellipse with vertices (–3, 1), (1, 1) and
co-vertices (–1, 2) and (–1, 0).
− + − =x hb
y ka
( ) ( )
1
2
2
2
2
− + − =x y( 2)
5
( 6)
13
1
2
2
2
2
− + − =x y( 2)
25
( 6)
169
1
2 2
= + + −⎛⎝⎜
⎞⎠⎟
M 2 2
2
,
19 ( 7)
2
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
+ =x y4 8
2 2
+ + − =x y x y25 4 50 32
2 2
Conic Sections 17
11.4 Hyperbolas
Hyperbolas share much in common with ellipses. While an ellipse is the
set of all points in the plane such that the sum of the distances from two
fixed points is constant, a hyperbola is the set of all points in the plane
such that the difference of the distances from two points is constant.
The two fixed points are called foci.
A hyperbola has two branches and two axes. The transverse axis connects
the two vertices of the hyperbola. The conjugate axis is perpendicular to
the transverse axis and intersects it at the midpoint of the two vertices,
or the center of the hyperbola. As the branches move away from each
center, they approach, but do not cross, one of two asymptotes.
-4 -3 -2 -1 1 2 3 4 5 6-5-6
-1
-2
-3
-4
-5
-6
-7
-8
2
3
4
5
6
7
8
1
(0, 8)
(0, 5)
(0, -5)
(0, -8)
Hyperbola with horizontal Hyperbola with vertical
transverse axis transverse axis
There are two forms of the standard equation of a hyperbola centered at
(0, 0), depending on whether the transverse axis is horizontal or vertical.
Standard Form of the Equation of a Hyperbola Centered at (0, 0)
Transverse Axis Horizontal Vertical
Equation
Foci (–c, 0), (c, 0) (0, –c), (0, c)
a, b, c c2
= a2
+ b2 c2
= a2
+ b2
Vertices (–a, 0), (a, 0) (0, –a), (0, a)
Asymptotes
Notice the many similarities between the equations and properties of an
ellipse centered at (0, 0) and a hyperbola centered at (0, 0). Do note,
however, the different relationship between a, b, and c. For an ellipse,
c2
= a2
– b2
and for a hyperbola, c2
= a2
+ b2
.
− =ya
xb
1
2
2
2
2
− =xa
yb
1
2
2
2
2
= = −y ab
x y ab
x,= = −y ba
x y ba
x,
Master Math: Algebra 218
When a hyperbola is written in standard form, you can identify the directionof the transverse axis by identifying which variable is not being subtracted. Ifx2 is not being subtracted, the transverse axis will be the x-axis or parallel tothe x-axis. If y2 is not being subtracted, the transverse axis will be the y-axisor parallel to the y-axis.
conjugate axis
asymptote
asymptote transverse axis
centervertexvertex focusfocus
conjugate axis
asymptote
asymptote
transverse axis
center
vertex
vertex
focus
focus
Conic Sections
You can write the equation of a hyperbola if you know the coordinates
of the vertices and the foci.
Example: Write the equation of the hyperbola.
Solution: The hyperbola has a horizontal
transverse axis and is centered at the
origin. The coordinates of the foci are
(–10, 0), and (10, 0), so c = 10. The
coordinates of the vertices are (–8, 0)
and (8, 0), so a = 8. Use the equation
c2
= a2
+ b2
to find the value of b2
.
Write the equation of the hyperbola using the standard equation
.
c2
= a2
+ b2
Relationship among a, b, and c.
10
2
= 8
2
+ b2
Substitute c = 10 and a = 8.
100 = 64 + b2
Simplify.
36 = b2
Subtract 64 from both sides of the equation.
Standard equation of hyperbola with horizontal
transverse axis.
Substitute a = 8 and 36 = b2
.
Simplify.
To graph a hyperbola, graph the vertices, foci, and asymptotes, and then
sketch the graph.
Example: Write the equation in standard form and
sketch a graph of the hyperbola.
Solution: Transform the equation into standard form. Identify a2
and
b2
, and solve to find a and b. Use the equation c2
= a2
+ b2
to find
c. Plot the foci, vertices, and asymptotes, and then sketch the graph.
− =xa
yb
1
2
2
2
2
− =xa
yb
1
2
2
2
2
− =x y8 36
1
2
2
2
− =x y64 36
1
2 2
− − =y x16 9 144 0
2 2
19
-10 -5 5 10
-5
5
(10, 0)(8, 0)(-8, 0)(-10, 0)
Add 144 to both sides of the equation.
Divide both sides of the equation by 144.
Since y2
is not being subtracted, the hyperbola has a vertical
transverse axis and the standard equation is .
a2
= 9 b2
= 16 c2
= a2
+ b2
a = 3 b = 4 c2
= 9 + 16
c2
= 25
c = 5
Foci: (0, –c), (0, c) → (0, –5), (0, 5)
Vertices: (0, –a), (0, a) → (0, –3), (0, 3)
Asymptotes: →
The graph of a hyperbola can be translated so that the center is (h, k).
Just as you have seen with other translated graphs, x – h and y – kreplace x and y.
− =y x9 16
1
2 2
− =ya
xb
1
2
2
2
2
= = −y x y x3
4
,
3
4
= = −y ab
x y ab
x,
− − =y x16 9 144 0
2 2
− =y x16 9 144
2 2
Master Math: Algebra 220
-10 -5 5 10
-5
-10
10
5(0, 5)
(0, -5)
(0, 3)
(0, -3)
y = - x34
y = x34
– = 1y2
9x2
16
Conic Sections
Standard Form of the Equation of a Hyperbola Centered at (h, k)
Transverse Axis Horizontal Vertical
Equation
Foci (h – c, k), (h + c, k) (h, k – c), (h, k + c)
a, b, c c2
= a2
+ b2 c2
= a2
+ b2
Vertices (h – a, k), (h + a, k) (h, k – a), (h, k + a)
Asymptotes Lines through (h, k) Lines through (h, k)
with slopes with slopes
Example: Write the equation in standard
form and sketch a graph of the hyperbola.
Solution: Transform the equation into standard form by completing
the square for x and y and then divide both sides of the equation by
a value that will make the equation equal to 1. Identify a2
and b2
and solve to find a and b. Use the equation c2
= a2
+ b2
to find c.
Plot the foci, vertices, and asymptotes, and sketch the graph.
Regroup and factor.
;
Factor.
Divide both sides by 36.
Since x2
is not being subtracted, the ellipse has a horizontal
transverse axis and the standard equation is .
+ − + =x x y y9 54 4 40 55
2 2
± ba
± ab
− − − =x ha
y kb
( ) ( )
1
2
2
2
2
− − − =y ka
x hb
( ) ( )
1
2
2
2
2
+ − + =x x y y9 54 4 40 55
2 2
+ − − =x x y y9( 6 ) 4( 10 ) 55
2 2
= −100−⎛⎝⎜
⎞⎠⎟
4
10
2
2
= 81
⎛⎝⎜
⎞⎠⎟
9
6
2
2
+ + − − + = + −x x y y9( 6 9) 4( 10 25) 55 81 100
2 2
+ − − =x y9( 3) 4( 5) 36
2 2
+ − − =x y( 3)
4
( 5)
9
1
2 2
− − − =x ha
y kb
( ) ( )
1
2
2
2
2
21
The center of the hyperbola is (–3, 5).
a2
= 4 b2
= 9 c2
= a2
+ b2
a = 2 b = 3 c2
= 4 + 9
c2
= 13
c ≈ 3.61
Foci: (h – c, k), (h + c, k)
(–3 – 3.61, 5), (–3 + 3.61, 5)
(–6.61, 5), (0.61, 5)
Vertices: (h – a, k), (h + a, k)
(–3 – 2, 5), (–3 + 2, 5)
(–5, 5), (–1, 5)
Asymptotes: Lines through (–3, 5) with slopes
Graph the center, foci, and vertices. Graph the asymptotes by
starting at the center and graphing a slope of by going up 3 and
to the right 2 and plotting another point.
Graph the other asymptote by starting at the center and graphing a
slope of by going down 3 and to the right 2 and plotting another
point. Sketch the hyperbola.
± 3
2
3
2
− 3
2
Master Math: Algebra 222
-5 -4 -3 -2 1 2-1-6-7-8-9
3
4
2
1
-1
-2
6
7
8
5
(-3, 5)
(-5, 5) (0.61, 5)(-1, 5)(-6.61, 5)
– = 1(x + 3)2
4
(y – 5)2
9
Conic Sections
Practice Exercises
11.13 Write the equation of the hyperbola if the coordinates of the
foci are (0, 13), (0, –13) and the vertices are (0, 5), (0, –5).
11.14 Write the equation in standard form and sketch the
graph of the hyperbola.
11.15 Write the equation in standard
form and sketch the graph of the hyperbola.
11.5 Parabolas
Recall from Chapter 5 that the graph of a quadratic function is a
parabola. In Chapter 5, we focused on parabolas that open vertically,
either up or down, and were functions.
General Form Vertex Form
Example
− =x y4 4
2 2
− + − =y x y x16 9 64 108 404
2 2
( )= − −y x 2 4
2
= + +y x x2 3
2
23
-3 -2 -1 1
4
5
6
3
2
1
Vertex: (-1, 2)
(-2, 3)(0, 3)
Line of Symmetry x = -1
-2 -1 1 2 3 4 5
3
2
1
-1
-2
-3
-4
5
6
4
y = x2
y = (x – 2)2 – 4
Line of symmetry x = h
Vertex x-coordinate = (h, k)
Substitute x and solve for y.
y-intercept (0, c)
= + + ≠y ax bx c a, 0
2 = − + ≠y a x h k a( ) , 0
2
= −x ba2
−ba2
From a geometric perspective, a parabola can also be defined as the set
of all points in a plane that are the same distance from a line called the
directrix and a point not on the line called the focus. The midpoint of the
line segment perpendicular to the directrix and passing through the focus
is the vertex of the parabola. The line that passes through the vertex and
the focus is the axis of symmetry. The axis of symmetry is perpendicular
to the directrix.
In this study of parabolas, we will include parabolas that open vertically
as well as parabolas that open horizontally.
Parabola that opens vertically Parabola that opens horizontally
You can transform the equation of a parabola to a form that will help
you readily identify the direction, focus, and directrix of the parabola.
Standard Equations of a Parabola with Vertex (h, k)
Opens vertically Opens horizontally
Direction (up or down) (left or right)
Equation
Focus (h, k + p) (h + p, k)
Axis of symmetry x = h y = kDirectrix y = k – p x = h – pIf p > 0 opens up opens right
If p < 0 opens down opens left
− = −y k p x h( ) 4 ( )
2− = −x h p y k( ) 4 ( )
2
Master Math: Algebra 224
axis of symmetry
focus
vertex
directrix
d1 = d2
d1
d2 axis of symmetry focusvertex
directrix d1 = d2
d1
d2
Conic Sections
Notice that an equation with an x that is squared will open vertically,
while an equation with a y that is squared will open horizontally.
Example: Find the vertex, focus, and directrix of the parabola
and sketch a graph of the parabola.
Solution: Since y is the squared term, the parabola opens horizon-
tally. Use the standard equation . Write the
equation in standard form and identify h, k, and p. Graph the vertex,
focus, and directrix. Substitute a few values for x and find y to
identify a few points on the graph. Sketch a graph of the parabola.
Standard form.
Given equation.
Multiply both sides of the equation by –8.
Transform to standard form.
p = –2; the parabola opens left
Vertex: (h, k) → (0, 0)
Focus: (h + p, k) → (0 + (–2), 0) = (–2, 0)
Directrix: x = h – p → x = 0 – (–2) → x = 2
y (x, y)
–8 –8 (–8, –8)
–4 –2 (–2, –4)
4 –2 (–2, 4)
8 –8 (–8, 8)
Notice from the graph to the right
that the parabola is not a function
because it fails the vertical line test.
− =y x8
2
− = −y k p x h( ) 4 ( )
2
− = −y k p x h( ) 4 ( )
2
− =y x8
2
= −y x8
2
− = − −y x( 0) 4( 2)( 0)
2
− =y
x8
2
25
-8 -6 -4 -2 2 4
-2
-4
-6
-8
6
8
4
2
(0, 0)(-2, 0)
(-2, 4)
(-8, 8)
(-8, -8)
(-2, -4)
x = 2y2 = -8x
Example: Find the vertex, focus, and directrix of the parabola
and sketch a graph of the parabola.
Solution: Since x is the squared term, the parabola opens vertically.
Use the standard equation . Write the equation
in the general form by completing the square and identify h, k. and
p. Graph the vertex, focus, and directrix. Substitute a few values
for y and find x to identify a few points on the graph. Sketch a
graph of the parabola.
Standard equation.
Given equation.
Multiply both sides of the
equation by 6.
Subtract 22 from both sides.
; add 4 to both sides.
Factor and simplify.
Factor.
Transform to standard form.
p = ; the parabola opens up
Vertex: (h, k) → (2, 3)
Focus: (h, k + p) →
Directrix: y = k – p →
= − +y x x1
6
( 4 22)
2
− = −x h p y k( ) 4 ( )
2
− = −x h p y k( ) 4 ( )
2
− + =x x y1
6
( 4 22)
2
− + =x x y4 22 6
2
− = −x x y4 6 22
2
⎛⎝⎜
⎞⎠⎟
=4
2
4
2
− + = − +x x y( 4 4) 6 22 4
2
− = −x y( 2) 6 18
2
− = −x y( 2) 6( 3)
2
− =⎛⎝⎜
⎞⎠⎟
−x y( 2) 4
3
2
( 3)
2
3
2
+⎛⎝⎜
⎞⎠⎟
→⎛⎝⎜
⎞⎠⎟
2,3
3
2
2,
9
2
= − → =y y3
3
2
3
2
Master Math: Algebra 226
Conic Sections
x y
–1
0
4
5
You can use what you know about the vertex, focus, and directrix to
write the equation of a parabola.
Example: The point (0, 7) is the focus of a parabola that has its vertex
at (2, 7). Find the equation of the parabola.
Solution: The vertex is (2, 7), so h = 2 and k = 7. The focus and
the vertex have a common value of 7, so the axis of symmetry is the
horizontal line y = 7. The parabola opens horizontally and the focus
is (h + p, k). Use the focus to identify p and write the equation of
the parabola using the general equation .
Focus: (h + p, k) = (0, 7)
So h + p = 0 and h = 2.
2 + p = 0
p = –2
− +x x1( 46
22)2
9
2
− − − +1
6
(( 1) 4( 1) 22)
2
11
3
− +1
6
(0 4(0) 22)
2
11
3
− +1
6
(4 4(4) 22)
2
9
2
− +1
6
(5 4(5) 22)
2
− = −y k p x h( ) 4 ( )
2
− = −y k p x h( ) 4 ( )
2
− = − −y x( 7) 4( 2)( 2)
2
− = − −y x( 7) 8( 2)
2
27
-6 -4 -2 2 4 6 8
6
8
10
4
2y = (2, 3)
(2, )92
32
Practice Exercises
11.16 Find the vertex, focus, and directrix of the parabola , and
sketch a graph.
11.17 Find the vertex, focus, and directrix of the parabola
.
11.18 Find the equation of the parabola if the vertex is (6, 2) and the
directrix is x = 4.
11.6 Identifying Conic Sections
Conic sections are formed by a plane intersecting a cone. The equation
for a conic section can be written in a form that identifies the center or
vertex (h, k). Circles, ellipses, hyperbolas, and parabolas are all examples
of conic sections.
= − +y x y12 8 8
2
r
(h, k)
b
a(h, k)
=y x4
2
a
b(h, k)
Master Math: Algebra 228
Circle − + − =x h y k r( ) ( )
2 2 2
Ellipse − + − = >x ha
y kb
a b( ) ( )
1,
2
2
2
2
− + − = >x hb
y ka
a b( ) ( )
1,
2
2
2
2
Conic Sections
Each conic section is a special case of the general form of a second-degree
equation , where at least one of the
coefficients A, B, or C is non-zero. If you can transform a second-degree
equation into one of the conic section equations above, you can identify
and graph the conic section.
(h, k)
slopes = ± ba
(h, k)slopes = ± ab
p
p (h, k)
pp(h, k)
+ + + + + =Ax Bxy Cy Dx Ey F 0
2 2
29
Hyperbola − − − =x ha
y kb
( ) ( )
1
2
2
2
2
− − − =y ka
x hb
( ) ( )
1
2
2
2
2
Parabola − = −x h p y k( ) 4 ( )
2
− = −y k p x h( ) 4 ( )
2
Master Math: Algebra 230
Example: Identify and describe the conic section given by the general
equation .
Solution: Complete the square for x and y and transform the
equation to match one of the preceding conic section equations.
Identify and describe the conic section.
Given equation.
Regroup.
Subtract 13 from both sides of
equation.
; ; Add.
Factor.
The conic section is a circle with center (2, 5) and radius 4.
Note in the example that if you divided both sides of the equation by 16,
you would have the equation , which would appear
to be an ellipse. However, a = b, so the conic section would be a circle,
or a special case of the ellipse.
Many second-degree equations are very difficult to factor into the conic
section equations. However, you can use the values of the coefficients
A, B, and C in the equation to deter-
mine the type of conic section that the equation represents.
Discriminant Theorem of Conic Sections
If at least one of the coefficients A, B, or C is non-zero, then the graph
of is a(n)
• ellipse if B2
– 4AC < 0 and A ≠ C
• circle if B2
– 4AC < 0 and A = C
• hyperbola if B2
– 4AC > 0
• parabola if B2
– 4AC = 0
Recall that b2
– 4ac is the discriminant of a quadratic equation.
+ − − + =x y x y4 10 13 0
2 2
+ − − + =x y x y4 10 13 0
2 2
− + − + =x x y y( 4 ) ( 10 ) 13 0
2 2
− + − = −x x y y( 4 ) ( 10 ) 13
2 2
= 25
−⎛⎝⎜
⎞⎠⎟
10
2
2
= 4
⎛⎝⎜
⎞⎠⎟
4
2
2
− + + − + = − + +x x y y( 4 4) ( 10 25) 13 4 25
2 2
− + − =x y( 2) ( 5) 16
2 2
− + − =x y( 2)
16
( 5)
16
1
2 2
+ + + + + =Ax Bxy Cy Dx Ey F 0
2 2
+ + + + + =Ax Bxy Cy Dx Ey F 0
2 2
Conic Sections 31
Example: Identify the conic section given by the equation
.
Solution: Transform the equation into the general form of a second-
degree equation and identify A, B, and C. Calculate B2
– 4AC and
identify the conic section.
Given equation.
Add to both
sides of the equation.
A = 6, B = 3, C = 1
= –15
and A ≠ C
The conic section is an ellipse.
Example: Identify the conic section given by the equation
.
Solution: Transform the equation into the general form of a second-
degree equation and identify A, B, and C. Calculate B2
– 4AC and
identify the conic section.
Given equation.
Add 5xy + 3x –
7y + 3 to both
sides.
A = 1, B = 5, C = –2
= 33
The conic section is a hyperbola.
+ + = − + −x x xy y y6 5 3 2 18
2 2
+ + = − + −x x xy y y6 5 3 2 18
2 2
+ + + − + =x xy y x y6 3 5 2 18 0
2 2 −y y2
2
− = −B AC4 3 4(6)(1)
2 2
− <B AC4 0
2
− = − − + −x y xy x y2 5 3 7 3
2 2
− = − − + −x y xy x y2 5 3 7 3
2 2
+ − + − + =x xy y x y5 2 3 7 3 0
2 2
− = − −B AC4 5 4(1)( 2)
2 2
− >B AC4 0
2
Practice Exercises
Identify each conic section.
11.19
11.20
11.7 Solve Quadratic Systems
A quadratic system is a system of equations where at least one equation
is quadratic. Solutions to a system can be represented by the points of
intersection of the graphs of the equations. An independent quadratic
system can have zero or up to four solutions.
Consider the following systems:
0 Solutions 1 Solution 2 Solutions 3 Solutions 4 Solutions
Solve by Graphing
One way to solve a quadratic system is by graphing. Graph each equa-
tion and locate the point or points of intersection.
Example: Solve the system by graphing.
Solution: The first equation is a circle centered at the origin with
radius 10. The second equation is a line. Solve the linear equation
for y and graph the line using the slope and the y-intercept.
Identify the points of intersection.
3x – y = 10
–y = –3x + 10 Subtract 3x from both sides of the equation.
y = 3x – 10 Multiply both sides of the equation by –1.
+ + = − +x y xy x y4 4 3 2 11
2 2
= − − + −y x x y3 4 2
2 2
+ =− =
⎧⎨⎩⎪
x yx y
100
3 10
2 2
Master Math: Algebra 232
Conic Sections 33
slope = 3, y-intercept = –10
The points of intersection on the graph are (6, 8), (0, –10).
Check: (6, 8)
x2
+ y2
= 100 3x – y = 10
6
2
+ 8
2
= 100 3(6) – 8 = 10
36 + 64 = 100 ¸ 18 – 8 = 10 ¸
Check: (0, –10)
x2
+ y2
= 100 3x – y = 10
0
2
+ (–10)
2
= 100 3(0) – (–10) = 10
0 + 100 = 100 ¸ 0 + 10 = 10 ¸The solutions to the system are (6, 8), (0, –10).
Identifying solutions from a graph can be inefficient and difficult. The
system may be challenging to transform into the standard form of a
conic section, or the intersection points may not be integer values.
Algebraic methods for solving a quadratic system can be more efficient.
Two algebraic methods are substitution and combinations.
-4 -2 2 4 6 8 10-6-8-10
-2
-4
-6
-8
-10
6
8
10
4
2
(6, 8)
(0, -10)
Solve by Substitution
Substitution is an effective method if one of the equations is linear.
Solve the linear equation for x or y and substitute the value for the
corresponding variable in the other equation to find the value of one
variable. Substitute that value into one of the equations and solve for the
remaining variable. Check your solutions by graphing or by substituting
each solution into both of the original equations.
Example: Solve the system by substitution.
Solution: Solve the linear equation for x or y and substitute the
value for the corresponding variable in the quadratic equation.
Solve for the remaining variable. Substitute the known value into
one of the equations to find the remaining variable.
x + y = –1
y = –x – 1 Subtract x from both sides.
Substitute –x – 1 for y in the equation .
Substitute y = –x – 1.
Simplify.
Add x to and subtract 3 from both sides.
(x – 3)(x – 2) = 0 Factor.
x = 3 or 2 Zero Product Property.
Substitute to find y values.
If x = 3: If x = 2:
3 + y = –1 2 + y = –1
y = – 4 y = –3
(3, –4) (2, –3)
− = ++ = −
⎧⎨⎩⎪
x yx y( 3) 4
1
2
− = +x y( 3) 4
2
− = +x y( 3) 4
2
− = − − +x x( 3) 1 4
2
− + = − +x x x6 9 3
2
− + =x x5 6 0
2
Master Math: Algebra 234
Conic Sections 35
Check: (3, –4)
x + y = –1
3 + –4 = –1 ¸
0 = 0 ¸Check (2, –3):
x + y = –1
2 + –3 = –1 ¸
1 = 1 ¸
The solutions to the system are (3, –4), (2, –3).
Solve by Combinations
If both equations in a quadratic system are quadratic, the best method
to solve is usually by combinations. Write each equation in the general
form . Multiply one or both equations
by numbers so that when you add the equations together, a variable will
cancel out. Solve that equation to find the value of the variable.
− = +x y( 3) 4
2
− = − +(3 3) 4 4
2
− = +x y( 3) 4
2
− = − +(2 3) 3 4
2
+ + + + + =Ax Bxy Cy Dx Ey F 0
2 2
-1 1 2 3 4 5
-1
-2
-3
-4
-5
2
1
(2, -3)
(3, -4)
Substitute that value back into one of the original equations to find the
value of the other variable. Check your solutions.
Example: Solve the system by combinations.
Solution: Transform each equation into
. Add the equations together
to eliminate a variable. Solve the resulting equation to find the
value(s) of one variable. Substitute each value back into one of
the original equations to find the corresponding value of the other
variable. Check every solution in both equations or check by
graphing.
Eq. 1: Eq. 2:
Equation 1.
Equation 2.
Add Equation 1 and Equation 2.
Divide both sides of the equation
by 2.
(x – 6)(x – 2) = 0 Factor.
x = 6 or 2 Zero Product Property.
Substitute to find y values.
If x = 6: If x = 2:
Substitute
value of x.
Simplify.
Simplify.
− + − =− − + = −
⎧⎨⎪
⎩⎪
x yx x y y( 4) ( 1) 13
8 2 20
2 2
2 2
+ + + + + =Ax Bxy Cy Dx Ey F 0
2 2
− + − =x y( 4) ( 1) 13
2 2 − − + = −x x y y8 2 20
2 2
− + + − + =x x y y8 16 2 1 13
2 2 − − + + =x y x y8 2 20 0
2 2
+ − − + =x y x y8 2 4 0
2 2
+ − − + =x y x y8 2 4 0
2 2
− − + + =x y x y8 2 20 0
2 2
− + =x x2 16 24 0
2
− + =x x8 12 0
2
− − + = −y y2 8(2) 2 20
2 2− − + = −y y6 8(6) 2 20
2 2
− − + = −y y36 48 2 20
2 − − + = −y y4 16 2 20
2
− + + =y y2 8 0
2− + + =y y2 8 0
2
Master Math: Algebra 236
Conic Sections 37
Multiply equation
by –1.
(y – 4)(y + 2) = 0 (y – 4)(y + 2) = 0 Factor.
y = 4 or y = –2 y = 4 or y = –2 Zero Product Property.
(6, 4), (6, –2) (2, 4), (2, –2)
You can check the solutions by substituting each solution into
both equations. The check is left to you.
The solutions are (6, 4), (6, –2), (2, 4), (2, –2).
Practice Exercises
Use the following system for 11.21 and 11.22
11.21 Solve the system above by graphing.
11.22 Solve the system above by substitution.
11.23 Solve the system by combinations:
= += +
⎧⎨⎩⎪
y xy x
5
3
2
= ++ − =
⎧⎨⎪
⎩⎪
y xx y
3
( 4) 1
2
2 2
− − =y y2 8 0
2− − =y y2 8 0
2
1 2 3 4 5 6 7 8
-1
-2
-3
3
4
5
2
1
(6, 4)(2, 4)
(6, -2)(2, -2)
Chapter 11 Answers to Exercises
11.1 ; = (–2, 5)
The midpoint of is (–2, 5).
11.2 ; ; ;
; ; ; ;
The coordinates of D are (–8, –3).
11.3 ; ;
;
The length is approximately 10.05 units.
11.4 ; ;
; Center (–5, –2), radius 1
11.5 h = 2, k = 1, r = 9; ; ;
= + − +⎛⎝⎜
⎞⎠⎟
M 1 ( 5)
2
,
7 3
2
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
AB
− =+ x
3
2
2
2− =+ +⎛
⎝⎜⎞⎠⎟
x y( 3,1)
2
2
,
5
2
2 2=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
− = y3
2
= + y2 5
2
=+ y
1
5
2
2− = x8
2
− = + x6 2
2
( ) ( )= − − + − − −d 2 8 5 ( 4)
2 2( ) ( )= − + −d x x y y2 1
2
2 1
2
= ≈d 101 10.05= − + −d ( 10) ( 1)
2 2
AB
+ + + =x y( 5) ( 2) 1
2 2− + − =x h y k r( ) ( )
2 2 2
− − + − − =x y( ( 5)) ( ( 2)) 1
2 2 2
-5 -1-3 -2-6 -4
-1
-2
-3(x + 5)2 + (y + 2)2 = 1
(-5, -2
(-5, -1)
(-5, -3)
(-4, -2)(-6, -2)
− + − =x y( 2) ( 1) 9
2 2 2− + − =x h y k r( ) ( )
2 2 2
− + − =x y( 2) ( 1) 81
2 2
Master Math: Algebra 238
Answers to Exercises 39
11.6 ; ; (h, k) = (1, –6)
The center of the circle is (1, –6). ;
; ; . The radius of the
circle is . ; ;
11.7 ; ;
;
;
11.8 The coordinates of the vertices are (0, 8) and (0, –8), so the
length of the major axis is 16. 2a = 16; a = 8; ; The foci
are (0, 5) and (0, –5), so c is 5. ; ;
; ; ; ;
11.9 2a = 10; a = 5; 2b = 8; b = 4; ; ;
11.10 ; . Since the denominator of x2
is larger,
the ellipse has a horizontal major axis and the standard equation
is .
; ; a ≈ 2.83; ; ; b ≈ 1.41; ;
; ; ; c ≈ 2.45
( ) ( )= − + −d x x y y2 1
2
2 1
2
=r 8= +r 4 4( ) ( )= − + − − −r 1 3 6 ( 8)
2 2
− + − − =x y( 1) ( ( 6)) 8
2 2
2
− + − =x h y k r( ) ( )
2 2 2
8
− + + =x y( 1) ( 6) 8
2 2
+ + + = −x x y y( 12 ) ( 18 ) 106
2 2+ + + = −x y x y12 18 106
2 2
+ + + + + = − +x x y y( 12 36) ( 18 81) 106 117
2 2
+ + + =x y( 6) ( 9) 11
2 2
2
+ + + =x y( 6) ( 9) 11
2 2
=a 64
2
= − b5 64
2 2= −c a b2 2 2
+ =x y39 64
1
2 2
+ =xb
ya
1
2
2
2
2
= b39
2− = −b39
2= − b25 64
2
+ =x y4 5
1
2
2
2
2
+ =xb
ya
1
2
2
2
2
+ =x y16 25
1
2 2
+ =x y8 2
1
2 2
+ =x y4 8
2 2
= − + − + −⎛⎝⎜
⎞⎠⎟
h k( , )
1 3
2
,
4 8
2
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
+ =xa
yb
1
2
2
2
2
= −c a b2 2 2=b 2=b 2
2=a 8=a 8
2
=c 6=c 6
2= −c 8 2
2
Master Math: Algebra 240
Foci: (–c, 0), (c, 0) → (–2.45, 0), (2.45, 0)
Vertices: (–a, 0), (a, 0) → (–2.83, 0), (2.83, 0)
Co-Vertices: (0, –b), (0, b) → (0, –1.41), (0, 1.41)
11.11 ; ;
;
; . Since the
denominator of y2
is larger, the ellipse has a vertical major axis
and the standard equation is . The center of
the ellipse is (–1, 4). a2
= 25; a = 5; b2
= 4; b = 2; ;
; ; c ≈ 4.58;
Foci: (h, k – c) = (–1, 4 – 4.58) = (–1, –0.58);
(h, k + c) = (–1, 4 + 4.58) = (–1, 8.58);
Vertices: (h, k – a) = (–1, 4 – 5) = (–1, –1);
(h, k + a) = (–1, 4 + 5) = (–1, 9);
Co-Vertices: (h – b, k) = (–1 – 2, 4) = (–3, 4);
(h + b, k) = (–1 + 2, 4) = (1, 4)
-1 31 2-2-3
1
2
-1
-2
x2 + 4y2 = 8(0, 1.41)
(0, -1.41)
(2.83, 0)(-2.45, 0) (2.45, 0)(-2.83, 0)
+ + − =x y x y25 4 50 32 11
2 2 + + − =x x y y25( 2 ) 4( 8 ) 11
2 2
+ + + − + = + +x x y y25( 2 1) 4( 8 16) 11 25 64
2 2
+ + − =x y( 1)
4
( 4)
25
1
2 2
+ + − =x y25( 1) 4( 4) 100
2 2
− + − =x hb
y ka
( ) ( )
1
2
2
2
2
= −c a b2 2 2
=c 21
2= −c 25 4
2
Answers to Exercises
11.12 ; M = (–1, 1); h = –1
and k = 1. Length of the horizontal axis = 1 – (–3) = 4; Length
of the vertical axis = 2 – 0 = 2; 2a = 4; a = 2; 2b = 2; b = 1
; ;
11.13 Foci: (0, 13), (0, –13), so c = 13; Vertices: (0, 5), (0, –5),
so a = 5; ; ; ; ;
; ;
11.14 ; ; Since x2
is not being subtracted, the
hyperbola has a horizontal transverse axis and the standard
equation is . a2
= 4; a = 2; b2
= 1; b = 1
= − + +⎛⎝⎜
⎞⎠⎟
M 3 1
2
,
1 1
2
=+ +⎛
⎝⎜⎞⎠⎟
Mx x y y
2
,
2
1 2 1 2
+ + − =x y( 1)
4
( 1)
1
1
2 2+ + − =x y( 1)
2
( 1)
1
1
2
2
2
2
− + − =x ha
y kb
( ) ( )
1
2
2
2
2
= b144
2= + b169 25
2= + b13 5
2 2 2= +c a b2 2 2
− =y x25 144
1
2 2
− =y x5 144
1
2
2
2
− =ya
xb
1
2
2
2
2
− =x y4 1
1
2 2
− =x y4 4
2 2
− =xa
yb
1
2
2
2
2
41
-1 1-2-3
9
8
7
6
5
4
3
2
1
-1
25x2 + 4y2 + 50x – 32y = 11
(-1, 9)
(-1, 8.58)
(-1, -0.58)
(1, 4)(-3, 4)
(-1, -1)
; c2
= 4 + 1; c2
= 5;
Foci: (–c, 0), (c, 0) → (–2.24, 0), (2.24, 0)
Vertices: (–a, 0), (a, 0) → (–2, 0), (2, 0)
Asymptotes: →
11.15 ; ;
;
;
; Since y2
is not being subtracted, the
ellipse has a vertical transverse axis and the standard equation is
. The center of the hyperbola is (–6, –2).
a2
= 9; a = 3; b2
= 16; b = 4; ; c2
= 9 + 16; c2
= 25;
c = 5; Foci: (h, k – c) = (–6, –2 – 5) = (–6, –7); (h, k + c) =
(–6, –2 + 5) = (–6, 3); Vertices: (h, k – a) = (–6, –2 – 3) = (–6, –5);
(h, k + a) = (–6, –2 + 3) = (–6, 1). Asymptotes: slope ± = ± .
+ − + =y x( 2)
9
( 6)
16
1
2 2
− − − =y ka
x hb
( ) ( )
1
2
2
2
2
= +c a b2 2 2
= ≈c 5 2.24= +c a b2 2 2
= = −y x y x1
2
,
1
2
= = −y ba
x y ba
x,
+ − + =y y x x16( 4 ) 9( 12 ) 404
2 2− + − =y x y x16 9 64 108 404
2 2
+ + − + + = −y y x x16( 4 4) 9( 12 36) 404 260
2 2
+ − + =y x16( 2) 9( 6) 144
2 2
ab
3
4
Master Math: Algebra 242
-1 31 2-2-3
2
1
-1
-2
y = - x y = xx2 – 4y2 = 4
(-2.24, 0) (-2, 0) (2, 0) (2.24, 0)
12
12
Answers to Exercises
11.16 Since x is the squared term, the parabola opens vertically.
; ; ;
p = 1; the parabola opens up; Vertex: (h, k) → (0, 0)
Focus: (h, k + p) → (0, 0 + 1) = (0, 1);
Directrix: y = k – p → y = 0 – 1 → y = –1
x y (x, y)
–4 4 (–4, 4)
–2 1 (–2, 1)
2 1 (2, 1)
4 4 (4, 4)
−( 4)
4
2
−( 2)
4
2
(2)
4
2
− = −x y( 0) 4(1)( 0)
2=y x4
2=y x4
2
− = −x h p y k( ) 4 ( )
2
x4
2
(4)
4
2
43
-6 2-2-4-8-10-12-14
4
2
-4
-2
-6
-8
16y2 – 9x2 + 64y – 108x = 404
(-6, 3)
(-6, 1)
(-6, -2)
(-6, -5)
(-6, -7)
-4 42-2
4
2
-2y = -1
(x – 0)2 = 4(1)(y – 0)
(0, 0)
(-2, 1) (2, 1)
(4, 4)(-4, 4)
(0, 1)
11.17 Since y is the squared term, the parabola opens horizontally.
; ; ;
; ; ;
; p = 3; the parabola opens to the right
Vertex: (h, k) → (–2, –4); Focus: (h + p, k) → (–2 + 3, –4) →
(1, –4); Directrix: x = h – p → x = –2 – 3 → x = –5
11.18 The vertex is (6, 2), and the directrix is x = 4, so the parabola opens
horizontally and to the right. h = 6, k = 2; x = h – p; 4 = 6 – p;
–2 = –p; 2 = p; ; ;
11.19 ;
A = 4, B = 4, C = 1; = 0; ;
The conic section is a parabola.
11.20 ; ;
; A = 1, B = 0, C = 1
= –4; and A = C.
The conic section is a circle.
11.21 The first equation is a line with a slope of 1 and a y-intercept of 5.
; ; ;
The second equation is a parabola with vertex (0, 3) that opens up.
x x2 + 3 f(x)
–2 (–2)
2
+ 3 7
–1 (–1)
2
+ 3 4
1 (1)
2
+ 3 4
2 (2)
2
+ 3 7
+ + + − + =x xy y x y0 3 4 2 0
2 2
− <B AC4 0
2− = −B AC4 0 4(1)(1)
2 2
− =y x3
2= +y x 3
2 − = −x y( 0) 3
2 ( )− =⎛⎝⎜
⎞⎠⎟
−x y( 0) 4
1
4
3
2
+ + − + − =x xy y x y4 4 3 2 11 0
2 2+ + = − +x y xy x y4 4 3 2 11
2 2
− =B AC4 0
2− = −B AC4 4 4(4)(1)
2 2
+ + − + =x y x y3 4 2 0
2 2= − − + −y x x y3 4 2
2 2
− = −y x( 2) 8( 6)
2
− = −y x( 2) 4(2)( 6)
2− = −y k p x h( ) 4 ( )
2
− − = − −y x( ( 4)) 4(3)( ( 2))
2
+ = +y x( 4) 12( 2)
2+ = +y x( 4) 12 24
2+ + = + +y y x8 16 12 8 16
2
+ = +y y x8 12 8
2= − +y x y12 8 8
2− = −y k p x h( ) 4 ( )
2
Master Math: Algebra 244
Answers to Exercises
The points of intersection appear to be (–1, 4) and (2, 7).
Check (–1, 4): y = x + 5; 4 = –1 + 5 ¸; ; ¸
Check (2, 7): y = x + 5; 7 = 2 + 5 ¸; ; ¸The solutions to the system are (–1, 4) and (2, 7).
11.22 ; ; ; ;
; x = 2 or x = –1. If x = 2: y = 2 + 5; y = 7;
(2, 7). If x = –1: y = –1 + 5; y = 4; (–1, 4). The graph from 10.21
verifies the solutions (2, 7) and (–1, 4).
11.23 Equation 1: ; ; ; Equation 2:
; ;
− + =y y7 12 0
2
− + − =x y 3 0
2− + =x y 3 0
2= +y x 3
2
+ − + =x y y8 15 0
2 2+ − + =x y y8 16 1
2 2+ − =x y( 4) 1
2 2
− + − =x y 3 0
2
+ − + =x y y8 15 0
2 2
= − +4 ( 1) 3
2= +y x 3
2
= +7 2 3
2= +y x 3
2
= − −x x0 2
2= − + −x x0 3 5
2+ = +x x5 3
2= +y x 3
2
= − +x x0 ( 2)( 1)
45
-1 1 2-2
7
6
5
4
3
2
1
y = x + 5
y = x2 + 3
(0, 3)
(1, 4)
(2, 7)
(-1, 4)
(-2, 7)
(y – 3)(y – 4) = 0; y = 3 or y = 4. If y = 3: ; 0 = x2
;
0 = x; (0, 3). If y = 4: ; 1 = x2
; x = 1 or x = –1;
(1, 4), (–1, 4). You can check the solutions by substituting each
solution into both equations. The check is left to you.
The solutions are (0, 3), (1, 4), and (–1, 4).
= +x3 3
2
= +x4 3
2
Master Math: Algebra 246
-1 1 2-2
6
7
5
4
3
2
1
y = x2 + 3
x2 + (y – 4)2 = 1 (0, 3)
(1, 4)(-1, 4)
